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Some vector calculus questions

  1. Oct 13, 2015 #1
    1. The problem statement, all variables and given/known data

    2. The shape of a hill is described by the height function h(x,y) = (2 + x2 + y4)-1/2

    (a) Find the gradient ∇h(x,y)

    (b) What is the maximum slope of the hill at the point r0 = i+j [or (x,y) = (1,1)]?

    (c) If you walk north-east (in the direction of the vector i+j) from the point r0, what
    is the slope of your path? Are you going uphill or downhill?

    (d) Find a vector in the direction of a path through the point r0 that remains at the
    same height.

    2. Relevant equations


    3. The attempt at a solution

    For parts a), b) and c) I think I'm okay.

    a) ∇h(x,y) = ([-x(2 + x2 + y4)-3/2], [-y3(2 + x2 + y4)-3/2])

    b) At point (1,1) just plug x and y into ∇h(x,y) to get (-1/8 , -1/4)

    c) Along path of vector (i+j) you will be going in a downhill direction since it is in a direction 'away' from the maximum slope found in the previous question. But is there a way I can mathematically prove this?

    d) This is the part I'm unsure of. It's been a while since I did work with vectors. I'm guessing I need to find a vector such that infinitesimal change in the x-y direction gives no change in height. So that would mean d(h(x,y)) = 0...? Since this would signify no change in the value of h(x,y). Overall I'm a bit lost.

    Thanks.
     
  2. jcsd
  3. Oct 13, 2015 #2

    SteamKing

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    You might want to recheck your calculation of ∂h / ∂y.
     
  4. Oct 13, 2015 #3
    Sorry, typo... should be -2y3 there.
     
  5. Oct 13, 2015 #4

    SteamKing

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    Good. Recalculate your results from that point on.
     
  6. Oct 13, 2015 #5
    No need. I'll get the same result. I've already done all of this on paper.The mistake you saw was purely a typing error.

    The only bit I'm stumped on is part d)
     
  7. Oct 13, 2015 #6

    Ray Vickson

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    If you go along the surface ##z = h(x,y)## a short distance from a point with ##x\,y## coordinates ##(x_0,y_0)## to a neighboring point with ##x\,y## coordinates ##(x_0+ \Delta x , y_0 + \Delta y)##, the change in "height" ##z## is given as
    [tex] \Delta z = h_x(x_0,y_0) \Delta x + h_y(x_0,y_0) \Delta y [/tex]
    to first order in the small step sizes ##\Delta x, \Delta y##. Here, ##h_x = \partial h/\partial x## and ##h_y = \partial h/ \partial y##. The equation above is standard Calculus II material.
     
  8. Oct 13, 2015 #7
    Sure, I get that, but how does it apply to part d) of the question?

    Thanks.
     
  9. Oct 13, 2015 #8

    Ray Vickson

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    PF Rules forbid us from telling you that.
     
  10. Oct 13, 2015 #9
    :oldeek: If it's not possible to get even a single nudge in the right direction, then it must mean that single nudge would give the whole answer, so I assume the whole thing must be far easier than I've been anticipating!

    For what it's worth, I've actually gone beyond even Calc IV at this point. I'm just rusty with vectors as I took a year out of university.

    I'll mull things over and try to see the glaringly obvious thing that I've missed.

    Thanks! :oldsmile:
     
  11. Oct 13, 2015 #10

    Ray Vickson

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    Hint: what is ##\Delta z## when you stay at the same level?
     
  12. Oct 14, 2015 #11
    Well it's 0, however I have no idea how that relates to finding a vector pointing in a direction such that its change in the z-direction is 0

    Intuitively it seems to be that a vector that is orthogonal to ∇h(x,y) will be at that point moving in a direction such that it has no change in z-position.

    ... which in the case of my given answer would be (1/4 i - 1/8 j ) ....?

    However I've done no calculus to get to that, which seems to go against the advice you're giving regarding ##\Delta z## etc. However I did work out ∇h(x,y) and the direction of steepest slope in the previous parts of the question, if that's what you were referring to.
     
  13. Oct 14, 2015 #12

    Ray Vickson

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    I don't get why you are having a problem. Saying that ##\Delta z = 0## is saying that ##z## (the height) does not change. And, since we already have a formula for ##\Delta z## in terms of ##\Delta x, \Delta y## and the partial derivatives of ##h(x,y)##, we have everything we need. Of course, that only determines the direction (the ratio of ##\Delta y## to ##\Delta x##) and not the absolute magnitudes of ##\Delta x, \Delta y##, so we are free to "scale" the results however we like.
     
  14. Oct 20, 2015 #13
    Sure but does ∇h(x,y) not determine the direction of x and y such that the resultant vector is pointing in the direction of the steepest slope at that point of the function? Whereas my challenge is to find the vector which has no steepness at that point. Hence me intuitively suggesting it must be orthogonal to the gradient vector. Was I wrong?

    Thanks
     
  15. Oct 20, 2015 #14

    Ray Vickson

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    Why are you arguing with me? I said exactly the same as what you have said, except that I gave reasons instead of just writing down a prescription without explanation.
     
  16. Oct 20, 2015 #15
    Oh wow no I wasn't arguing, sorry if it came across that way. It was just the first line of your previous post seemed to infer that I was still wrong overall, so I remained confused.

    If all is good then that's fine, thanks.
     
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